在转换一个向量的元素时连接两个向量的最佳方法是什么?

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问题描述

假设我有

std::vectorvec1 {/* 填充了 T1 的 */};std::vectorvec2 {/* 填充 T2 的 */};

和一些函数 T1 f(T2) 当然可以是一个 lambda.在将 f 应用于 vec2T2 的同时连接 vec1 和 vec2 的最佳方法是什么?/代码>?

显而易见的解决方案是std::transform，即

vec1.reserve(vec1.size() + vec2.size());std::transform(vec2.begin(), vec2.end(), std::back_inserter(vec1), f);

但我说这不是最佳，因为std::back_inserter必须对每个插入的元素进行不必要的容量检查.什么是最佳的，就像

vec1.insert(vec1.end(), vec2.begin(), vec2.end(), f);

只需一次容量检查就可以逃脱.遗憾的是，这不是有效的 C++.本质上，这与 std::vector::insert 最适合向量连接的原因相同，请参阅 this 问题和 this 问题以进一步讨论这一点.

所以:

std::transform 是使用 STL 的最佳方法吗?
如果是这样，我们可以做得更好吗?
上述 insert 函数被排除在 STL 之外是否有充分的理由?

更新

我已经尝试验证多次容量检查是否确实有任何明显的成本.为此，我基本上只是将 id 函数 (f(x) = x) 传递给 std::transform 和 push_back 中讨论的方法答案.完整代码为:

#include #include #include #include #include #include <chrono>#include #include 使用 std::size_t;std::vectorgenerate_random_ints(size_t n){std::default_random_engine 生成器；自动种子1 = std::chrono::system_clock::now().time_since_epoch().count();generator.seed((无符号)seed1);std::uniform_int_distribution制服 {};std::vectorv(n);std::generate_n(v.begin(), n, [&] () { return uniform(generator); });返回 v;}模板 <typename D=std::chrono::nanoseconds, typename F>D 基准(F f，无符号 num_tests){D 总{0}；for (unsigned i = 0; i (结束 - 开始)；}返回 D {total/num_tests};}模板 void std_insert(std::vector vec1, const std::vector &vec2){vec1.insert(vec1.end(), vec2.begin(), vec2.end());}template void push_back_concat(std::vector vec1, const std::vector &vec2, UnaryOperation op){vec1.reserve(vec1.size() + vec2.size());for (const auto& x : vec2) {vec1.push_back(op(x));}}template void transform_concat(std::vector vec1, const std::vector &vec2, UnaryOperation op){vec1.reserve(vec1.size() + vec2.size());std::transform(vec2.begin(), vec2.end(), std::back_inserter(vec1), op);}int main(int argc, char **argv){无符号 num_tests {1000};size_t vec1_size {10000000};size_t vec2_size {10000000};自动 vec1 = generate_random_ints(vec1_size);自动 vec2 = generate_random_ints(vec1_size);auto f_std_insert = [&vec1, &vec2] () {std_insert(vec1, vec2);};auto f_push_back_id = [&vec1, &vec2] () {push_back_concat(vec1, vec2, [] (int i) { return i; });};auto f_transform_id = [&vec1, &vec2] () {transform_concat(vec1, vec2, [] (int i) { return i; });};auto std_insert_time = benchmark<std::chrono::milliseconds>(f_std_insert, num_tests).count();auto push_back_id_time = benchmark(f_push_back_id, num_tests).count();auto transform_id_time = benchmark(f_transform_id, num_tests).count();std::cout <<标准插入:" <<std_insert_time <<毫秒"<

编译:

g++ vector_insert_demo.cpp -std=c++11 -O3 -o vector_insert_demo

输出:

std_insert:44mspush_back_id:61 毫秒变换 ID:61 毫秒

编译器将内联 lambda，因此可以安全地降低成本.除非其他人对这些结果有一个可行的解释(或愿意检查组件)，否则我认为可以合理地得出多项容量检查的成本很高的结论.

解决方案

UPDATE:性能差异是由于 reserve() 调用造成的，至少在 libstdc++ 中，它使容量恰好是您请求的内容，而不是使用指数增长因子.

我做了一些计时测试，结果很有趣.使用 std::vector::insert 和 boost::transform_iterator 是我发现的最快方法:

版本 1:

voidappendTransformed1(std::vector&vec1,const std::vector&vec2){auto v2begin = boost::make_transform_iterator(vec2.begin(),f);auto v2end = boost::make_transform_iterator(vec2.end(),f);vec1.insert(vec1.end(),v2begin,v2end);}

版本 2:

voidappendTransformed2(std::vector&vec1,const std::vector&vec2){vec1.reserve(vec1.size()+vec2.size());对于(自动 x:vec2){vec1.push_back(f(x));}}

版本 3:

voidappendTransformed3(std::vector&vec1,const std::vector&vec2){vec1.reserve(vec1.size()+vec2.size());std::transform(vec2.begin(),vec2.end(),std::inserter(vec1,vec1.end()),f);}

时间:

版本 1:0.59s版本 2:8.22s版本 3:8.42smain.cpp:
#include #include <cassert>#include <chrono>#include #include #include #include #include "appendtransformed.hpp"使用 std::cerr;模板<类型名称引擎>静态 std::vectorrandomInts(Engine &engine,size_t n){自动分配 = std::uniform_int_distribution(0,999);自动生成器 = [&]{返回分布(引擎)；}；auto vec = std::vector();std::generate_n(std::inserter(vec,vec.end()),n,generator);返回 vec;}模板<类型名称引擎>静态 std::vectorrandomFloats(Engine &engine,size_t n){自动分配 = std::uniform_real_distribution(0,1000);自动生成器 = [&]{返回分布(引擎)；}；auto vec = std::vector();std::generate_n(std::inserter(vec,vec.end()),n,generator);返回 vec;}静态自动appendTransformedFunction(int 版本) ->void(*)(std::vector&,const std::vector &){开关(版本){情况1:返回appendTransformed1；情况 2:返回 appendTransformed2；情况 3:返回 appendTransformed3；默认:cerr<
编译器:g++ 4.9.1
选项:-std=c++11 -O2
Suppose I have
std::vector<T1> vec1 {/* filled with T1's */};
std::vector<T2> vec2 {/* filled with T2's */};
and some function T1 f(T2) which could of course be a lambda. What is the optimal way to concatenate vec1 and vec2 whilst applying f to each T2 in vec2?
The apparently obvious solution is std::transform, i.e.
vec1.reserve(vec1.size() + vec2.size());
std::transform(vec2.begin(), vec2.end(), std::back_inserter(vec1), f);
but I say this is not optimal as std::back_inserter must make an unnecessary capacity check on each inserted element. What would be optimal is something like
vec1.insert(vec1.end(), vec2.begin(), vec2.end(), f);
which could get away with a single capacity check. Sadly this is not valid C++. Essentially this is the same reason why std::vector::insert is optimal for vector concatenation, see this question and the comments in this question for further discussion on this point.
So:
Is std::transform the optimal method using the STL?
If so, can we do better?
Is there a good reason why the insert function described above was left out of the STL?
UPDATE
I've had a go at verifying if the multiple capacity checks do have any noticeable cost. To do this I basically just pass the id function (f(x) = x) to the std::transform and push_back methods discussed in the answers. The full code is:
#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>
#include <cstdint>
#include <chrono>
#include <numeric>
#include <random>

using std::size_t;

std::vector<int> generate_random_ints(size_t n)
{
    std::default_random_engine generator;
    auto seed1 = std::chrono::system_clock::now().time_since_epoch().count();
    generator.seed((unsigned) seed1);
    std::uniform_int_distribution<int> uniform {};
    std::vector<int> v(n);
    std::generate_n(v.begin(), n, [&] () { return uniform(generator); });
    return v;
}

template <typename D=std::chrono::nanoseconds, typename F>
D benchmark(F f, unsigned num_tests)
{
    D total {0};
    for (unsigned i = 0; i < num_tests; ++i) {
        auto start = std::chrono::system_clock::now();
        f();
        auto end = std::chrono::system_clock::now();
        total += std::chrono::duration_cast<D>(end - start);
    }
    return D {total / num_tests};
}

template <typename T>
void std_insert(std::vector<T> vec1, const std::vector<T> &vec2)
{
    vec1.insert(vec1.end(), vec2.begin(), vec2.end());
}

template <typename T1, typename T2, typename UnaryOperation>
void push_back_concat(std::vector<T1> vec1, const std::vector<T2> &vec2, UnaryOperation op)
{
    vec1.reserve(vec1.size() + vec2.size());
    for (const auto& x : vec2) {
        vec1.push_back(op(x));
    }
}

template <typename T1, typename T2, typename UnaryOperation>
void transform_concat(std::vector<T1> vec1, const std::vector<T2> &vec2, UnaryOperation op)
{
    vec1.reserve(vec1.size() + vec2.size());
    std::transform(vec2.begin(), vec2.end(), std::back_inserter(vec1), op);
}

int main(int argc, char **argv)
{
    unsigned num_tests {1000};
    size_t vec1_size {10000000};
    size_t vec2_size {10000000};

    auto vec1 = generate_random_ints(vec1_size);
    auto vec2 = generate_random_ints(vec1_size);

    auto f_std_insert = [&vec1, &vec2] () {
        std_insert(vec1, vec2);
    };
    auto f_push_back_id = [&vec1, &vec2] () {
        push_back_concat(vec1, vec2, [] (int i) { return i; });
    };
    auto f_transform_id = [&vec1, &vec2] () {
        transform_concat(vec1, vec2, [] (int i) { return i; });
    };

    auto std_insert_time   = benchmark<std::chrono::milliseconds>(f_std_insert, num_tests).count();
    auto push_back_id_time = benchmark<std::chrono::milliseconds>(f_push_back_id, num_tests).count();
    auto transform_id_time = benchmark<std::chrono::milliseconds>(f_transform_id, num_tests).count();

    std::cout << "std_insert: " << std_insert_time << "ms" << std::endl;
    std::cout << "push_back_id: " << push_back_id_time << "ms" << std::endl;
    std::cout << "transform_id: " << transform_id_time << "ms" << std::endl;

    return 0;
}
Compiled with:
g++ vector_insert_demo.cpp -std=c++11 -O3 -o vector_insert_demo
Output:
std_insert: 44ms
push_back_id: 61ms
transform_id: 61ms
The compiler will have inlined the lambda, so that cost can be safely be discounted. Unless anyone else has a viable explanation for these results (or is willing to check the assembly), I think it's reasonable to conclude there is a noticeable cost of the multiple capacity checks.
 解决方案 
UPDATE: The performance difference is due to the reserve() calls, which, in libstdc++ at least, make the capacity be exactly what you request instead of using the exponential growth factor.
I did some timing tests, with interesting results.  Using std::vector::insert along with boost::transform_iterator was the fastest way I found by a large margin:
Version 1:
void
  appendTransformed1(
    std::vector<int> &vec1,
    const std::vector<float> &vec2
  )
{
  auto v2begin = boost::make_transform_iterator(vec2.begin(),f);
  auto v2end   = boost::make_transform_iterator(vec2.end(),f);
  vec1.insert(vec1.end(),v2begin,v2end);
}
Version 2:
void
  appendTransformed2(
    std::vector<int> &vec1,
    const std::vector<float> &vec2
  )
{
  vec1.reserve(vec1.size()+vec2.size());
  for (auto x : vec2) {
    vec1.push_back(f(x));
  }
}
Version 3:
void
  appendTransformed3(
    std::vector<int> &vec1,
    const std::vector<float> &vec2
  )
{
  vec1.reserve(vec1.size()+vec2.size());
  std::transform(vec2.begin(),vec2.end(),std::inserter(vec1,vec1.end()),f);
}
Timing:
    Version 1: 0.59s
    Version 2: 8.22s
    Version 3: 8.42s
main.cpp:
#include <algorithm>
#include <cassert>
#include <chrono>
#include <iterator>
#include <iostream>
#include <random>
#include <vector>
#include "appendtransformed.hpp"

using std::cerr;

template <typename Engine>
static std::vector<int> randomInts(Engine &engine,size_t n)
{
  auto distribution = std::uniform_int_distribution<int>(0,999);
  auto generator = [&]{return distribution(engine);};
  auto vec = std::vector<int>();
  std::generate_n(std::inserter(vec,vec.end()),n,generator);
  return vec;
}

template <typename Engine>
static std::vector<float> randomFloats(Engine &engine,size_t n)
{
  auto distribution = std::uniform_real_distribution<float>(0,1000);
  auto generator = [&]{return distribution(engine);};
  auto vec = std::vector<float>();
  std::generate_n(std::inserter(vec,vec.end()),n,generator);
  return vec;
}

static auto
  appendTransformedFunction(int version) ->
    void(*)(std::vector<int>&,const std::vector<float> &)
{
  switch (version) {
    case 1: return appendTransformed1;
    case 2: return appendTransformed2;
    case 3: return appendTransformed3;
    default:
      cerr << "Unknown version: " << version << "\n";
      exit(EXIT_FAILURE);
  }

  return 0;
}

int main(int argc,char **argv)
{
  if (argc!=2) {
    cerr << "Usage: appendtest (1|2|3)\n";
    exit(EXIT_FAILURE);
  }
  auto version = atoi(argv[1]);
  auto engine = std::default_random_engine();
  auto vec1_size = 1000000u;
  auto vec2_size = 1000000u;
  auto count = 100;
  auto vec1 = randomInts(engine,vec1_size);
  auto vec2 = randomFloats(engine,vec2_size);
  namespace chrono = std::chrono;
  using chrono::system_clock;
  auto appendTransformed = appendTransformedFunction(version);
  auto start_time = system_clock::now();
  for (auto i=0; i!=count; ++i) {
    appendTransformed(vec1,vec2);
  }
  auto end_time = system_clock::now();
  assert(vec1.size() == vec1_size+count*vec2_size);
  auto sum = std::accumulate(vec1.begin(),vec1.end(),0u);
  auto elapsed_seconds = chrono::duration<float>(end_time-start_time).count();

  cerr << "Using version " << version << ":\n";
  cerr << "  sum=" << sum << "\n";
  cerr << "  elapsed: " << elapsed_seconds << "s\n";
}
Compiler: g++ 4.9.1
Options: -std=c++11 -O2
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